Question

if vertices of a triangle are (0,4), (4, 1) and
(7,5), then its perimeter is -
(A) 5(2+5) (B) 2 3
(C) 5(2 + √2) (D) 2/5​

Answer 1

Answer:

please mark it as a brainliest answer

Answer 2

Answer:

Perimeter of triangle = 5(2 + $\sqrt{2}$)

Step-by-step explanation:

Vertex of triangle-

A = (0,4)

B = (4,1)

C = (7,5)

We know that-

Distance between X(a,b) and Y(c,d) is-

XY = $\sqrt{(a-c)^{2}+(b-d)^{2}}$

So distance between A and B is-

AB = $\sqrt{(0-4)^{2}+(4-1)^{2}}$

     = $\sqrt{4^{2}+3^{2}}$

     = $\sqrt{16+9}$

     = $\sqrt{25}$

AB = 5

Distance between B and C is-

BC = $\sqrt{(4-7)^{2}+(1-5)^{2}}$

     = $\sqrt{(-3)^{2}+(-4)^{2}}$

     = $\sqrt{9+16}$

     = $\sqrt{25}$

BC = 5

Distance between C and A is-

CA = $\sqrt{(7-0)^{2}+(5-4)^{2}}$

     = $\sqrt{7^{2}+1^{2}}$

     = $\sqrt{49+1}$

     = $\sqrt{50}$

CA = $5\sqrt{2}$

Perimeter of triangle = Ab + BC + CA

                                  = 5 + 5 + $5\sqrt{2}$

                                  =  5(1 + 1 + $\sqrt{2}$)

Perimeter of triangle = 5(2 + $\sqrt{2}$)